CHEMISTRY NOTES
COLLISION THEORY
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For a chemical reaction to occur, the reacting molecules must collide with each other. How fast a reaction takes place
is very much dependent on the number of collisions that occur. Therefore:
number of collisions
rate ∝
seconds
Looking at the reaction: A + B → product
Doubling the concentrations of A and B would double the number of collisions. Therefore, the rate law is
rate = k[A][B]
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Looking at the reaction: 2A → product
Doubling the concentration of A would cause 4 times the number of collisions. Therefore, the rate law is
rate = k[A]2
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According to this theory, the exponents in the rate law is equal to the coefficient in the balanced equation.
o Some examples what this theory gives us:
2A + B → products
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rate = k[A]2[B]
D + 3E + 2F → products

rate = k[D][E]3[F]2
However, not all collisions are effective and lead to a reaction. In order to be effective, the following must occur:
o high enough frequency of collisions
o correct orientation
o enough energy to get close enough for change to occur in the bonding
Since the collision theory does not differentiate between effective and ineffective collisions we cannot use this theory
on its own to determine the rate law for a reaction. The ONLY way to determine a rate law is to do an experiment and
measure the reaction rates vs. concentrations to generate data like we worked on yesterday. The following section
will show where we can use the collision theory.
REACTION MECHANISMS
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The overall balanced equation for a reaction represents the net chemical change that occurs as the reaction proceeds
to completion. This does NOT mean that all the reactants come together simultaneously to undergo a change that
gives the products. The net change usually represents the sum of a series of simpler reactions—elementary
processes. The sequence of elementary processes that lead to the products is called the reaction mechanism.
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Looking at the reaction: 2NO + 2H2 → 2H2O + N2
It is possible that this reaction could take place in just one step. For this to happen both NO molecules and both H 2
molecules must come together at the same time, be lined up just right and have enough energy to produce the
products. Possible, but highly unlikely, especially if the reaction is somewhat quick. It is more likely that the reaction
will take place in a series of steps. The following is a possible reaction mechanism for the given reaction.
overall given reaction
2NO + 2H2 → 2H2O + N2
possible reaction mechanism
(elementary processes)
2NO → N2O2
N2O2 + H2 → N2O + H2O
N2O + H2 → N2 + H2O
Note that in each of the elementary processes there are only 2 particles reacting. The probability of effective
collisions occurring in each of these steps is quite high.
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It must be noted that reaction mechanisms are just proposals for how a given reaction may occur—we cannot actually
see the steps taking place. However, experimental data can be used to support such mechanisms. Are there other
possible mechanisms for this reaction? Yes there probably are.
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Note there are several substances in the mechanism that are not in the overall equation. These substances end up
cancelling out in the mechanism to give the overall reaction. These substances, like the N2O2 in step 1, produced in
one elementary process and later used as a reactant in another elementary process are called “intermediates”. The
intermediates in this reaction are: N2O2 and N2O.
2NO → N2O2
possible reaction mechanism
N2O2 + H2 → N2O + H2O
(elementary processes)
N2O + H2 → N2 + H2O
overall given reaction

2NO + 2H2 → 2H2O + N2
The different steps of the mechanism will take place at different speeds. How quickly an overall reaction can take
place will depend on the slowest elementary process—the reaction can only be as fast as its slowest step. The
slowest step in the mechanism is called the “rate determining step”. Using the collision theory and the rate
determining step, we can determine a rate law for the overall reaction.
o
If the first step, 2NO → N2O2, is the rate determining step, then according to the collision theory, the rate law for
this step is:
rate = k[NO]2
If this is the rate determining step then this will be the rate law for the entire overall reaction.
o
If the second step, N2O2 + H2 → N2O + H2O, is the rate determining step, then according to the collision theory,
the rate law for this step is:
rate = k[N2O2][H2]
However, this rate law cannot be the rate law for the overall reaction because N 2O2 is NOT in the overall reaction
—N2O2 is an intermediate. To take care of this problem you have to realize that the N 2O2 can only react as fast
as it is made in the first step. So, we can substitute the rate law for the first step into the rate law for the second
step.
rate = k[NO]2
rate = k[N2O2][H2]
By substituting k[NO]2 into the second rate law, we get the following.
rate = (k·k)[NO]2[H2]
Since “k” is a constant (and they are different values in the different steps but for simplicity purposes I am not
going to differentiate between the two), multiplying the two k’s will give you another k. So the rate law for the
entire reaction will therefore be the following.
rate = k[NO]2[H2]
o
If the third step, N2O + H2 → N2 + H2O, is the rate determining step, then according to the collision theory, the rate
law for this step is:
rate = k[N2O][H2]
Again we have the problem of having an intermediate in the rate law. So we need to substitute the rate law of the
second step into this one since N2O is only able to react as fast as it is made in the second step.
rate = k[NO]2[H2]
rate = k[N2O][H2]
This gives us:
rate = (k·k)[NO]2[H2][H2]
This simplifies to:
rate = k[NO]2[H2]2
To summarize:
if the rate determining step is the:
then the rate law for the overall reaction is:
1st
rate = k[NO]2
step
2nd step
rate = k[NO]2[H2]
3rd step
rate = k[NO]2[H2]2
By doing an experiment to determine the rate law for the reaction, we can determine which step is the rate
determining step.
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OK, this seems to be a LONG and drawn out process. The good news is there is a shortcut in doing this and all you
have to do is count. To explain this, I will use the same reaction and mechanism and apply the shortcut.
o
overall given reaction
2NO + 2H2 → 2H2O + N2
possible reaction mechanism
2NO → N2O2
N2O2 + H2 → N2O + H2O
N2O + H2 → N2 + H2O
If the first step, 2NO → N2O2, is the rate determining step, all you need to do is count any reactant that is in the
overall equation and that number is the order of reaction (the exponent) for that substance.
2NO → N2O2
Since there are 2 NO’s, then the rate law for the overall reaction is: rate = k[NO] 2
o
If the second step, N2O2 + H2 → N2O + H2O, is the rate determining step, starting with the 2nd step and working
back to the beginning, count any reactant that is in the overall equation to get the order of reaction and the rate
law.
2NO → N2O2
N2O2 + H2 → N2O + H2O
Doing this, we count 1 H2 and 2 NO’s, so the rate law for the overall reaction is: rate = k[NO]2[H2]
o
If the third step, N2O + H2 → N2 + H2O, is the rate determining step, starting with the 3rd step and working back to
the beginning, count any reactant that is in the overall equation to get the order of reaction and the rate law.
2NO → N2O2
N2O2 + H2 → N2O + H2O
N2O + H2 → N2 + H2O
Doing this, we count 2 H2’s and 2 NO’s, so the rate law for the overall reaction is: rate = k[NO] 2[H2]2
SOME EXAMPLES:
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For the following mechanism and overall reaction, determine the rate law for the overall reaction if the 2 nd step is the
rate determining step.
overall given reaction
2H2O2 → 2H2O + O2
H2O2 → 2OH
reaction mechanism
H2O2 + OH → H2O + HO2
HO2 + OH → H2O + O2
answer: rate = k[H2O2]2
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For the following mechanism and overall reaction, determine the rate law for the overall reaction if the 3rd step is the
rate determining step.
overall given reaction
C4H9Br + 2H2O → C4H9OH + Brˉ + H3O+
C4H9Br → C4H9+ + Brˉ
reaction mechanism
C4H9+ + H2O → C4H9OH2+
C4H9OH2+ + H2O → C4H9OH + H3O+
answer: rate = k[C4H9Br][H2O]2
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For the following mechanism and overall reaction, determine the rate law for the overall reaction if the 1st step is the
rate determining step.
overall given reaction
3/2 O2 → O3
NO + ½ O2 → NO2
reaction mechanism
NO2 → NO + O
O2 + O → O3
answer: rate =
o
o
k[O2]1/2
What are the intermediates in this mechanism? NO2 and O
There is another substance here that is not in the overall equation: NO
This is a catalyst. You can tell if you have a catalyst if it first appears as a reactant in the mechanism and later is
found as a product.
intermediates → found first as a product and later as a reactant
catalyst → found first as a reactant and later as a product
In either case, these substances cancel out of the mechanism and is not in the rate law for the reaction.